$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be

  • A
    $_{7}N^{14}$
  • B
    $_{7}N^{13}$
  • C
    $_{6}C^{12}$
  • D
    $_{7}N^{15}$

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